lim n → ∞ 1 3 . 7 + 1 7 . 11 + 1 11 . 15 + … + ( n   terms ) =

The series is given as,

$=\underset{n\to \infty }{\lim }\left(\frac{1}{3·7}+\frac{1}{7·11}+\frac{1}{11·15}+\dots +\left(n \text{terms}\right)\right)$

$=\underset{n\to \infty }{\lim }\frac{1}{4}\left(\frac{4}{3·7}+\frac{4}{7·11}+\frac{4}{11·15}\dots \right)$

$=\underset{n\to \infty }{\lim }\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\dots \frac{1}{n}+\frac{1}{n}-\frac{1}{n+4}\right)$

$=\underset{n\to \infty }{\lim }\frac{1}{4}\left(\frac{1}{3}-\frac{1}{n+4}\right)$

$=\underset{n\to \infty }{\lim }\frac{1}{12}-\underset{n\to \infty }{\lim }\frac{1}{4(n+4)}$

$=\frac{1}{12}-0=\frac{1}{12}$